Related Rates: Cone

Setup - Linda is bored and decides to pour an entire container of salt into a pile on the kitchen floor. She pours $$3$$ cubic inches of salt per second into a conical pile whose height is always two-thirds of its radius.

Question - How fast is the radius of the conical salt pile changing when the radius of the pile is $$2$$ inches?

So I don't really know what to do here. Could someone walk me through it? I'm assuming you use the volume and surface area formula of a cone.

First, let $$t$$ be time in seconds, $$r(t)$$ be the radius in inches at time $$t$$ seconds, $$h(t)$$ be the height in inches at time $$t$$ seconds, and $$V(t)$$ be the volume in cubic inches at time $$t$$ seconds, and let $$t_0$$ be the time at which the radius is $$2$$ inches.

What do we know? The volume is increasing at a rate of $$3$$ cubic inches per second, so $$\frac{\text dV}{\text dt}(t) = 3$$ for all $$t$$. Height is two-thirds the radius, so $$h(t) = \frac{2}{3}r(t)$$ for all $$t$$. The radius is $$2$$ inches at time $$t_0$$, so $$r(t_0) = 2$$

What do we want? We are asked how fast the radius is changing (with respect to time) at the instant where the radius is $$2$$ inches, so we want to find $$\frac{\text dr}{\text dt}(t_0)$$.

What do we need? We have a relationship between $$h$$ and $$r$$, but in terms of rates, we only have information about $$\frac{\text dV}{\text dt}$$. So, we need some relationship between $$V$$ and $$r$$, $$h$$, or both. Since the shape is a cone, we can can use the formula for the volume of a cone, $$V(t) = \frac{1}{3}\pi h(t)r(t)^2$$ Substituting our relationship between $$r$$ and $$h$$, we get $$V(t) = \frac{2}{9}\pi r(t)^3$$ Differentiating this gives us $$\frac{\text dV}{\text dt}(t) = \frac{2}{3}\pi r(t)^2\frac{\text dr}{\text dt}(t)$$

Now that we have all the information, we just need to put it together. Plugging in $$t=t_0$$ will give us the relationship when the radius is $$2$$ inches, $$\frac{\text dV}{\text dt}(t_0) = \frac{2}{3}\pi r(t_0)^2\frac{\text dr}{\text dt}(t_0)$$ However, we know that $$\frac{\text dV}{\text dt}(t) = 3$$ for all $$t$$, and in particular at $$t=t_0$$, so we have $$\frac{\text dV}{\text dt}(t_0) = 3$$ We also know that $$r(t_0) = 2$$ by definition of $$t_0$$. This means that our equation reduces to $$3 = \frac{2}{3}\pi\cdot 2^2 \frac{\text dr}{\text dt}(t_0)$$ which we can solve to determine the value $$\frac{\text dr}{\text dt}(t_0) = \frac{9}{8}\pi$$

In terms of the original problem, $$\frac{\text dr}{\text dt}(t_0) = \frac{9}{8}\pi$$ can be interpreted as the salt pile's radius increasing at a rate of $$\frac{9}{8}\pi$$ inches per second when the radius is $$2$$ inches.

We only have to use the formula for volume.

We know two things: $$V=\frac13\pi r^2h$$ and $$h=\frac23r$$.

Let's sub this in to get the volume, $$V$$, as a function of the radius, $$r$$.

$$V(r) = \frac29\pi r^3$$

Next, we derive this function with respect to time, $$t$$.

$$\frac{dV}{dt} = \frac23\pi r^2 \frac{dr}{dt}$$

We were given a few values in the setup: $$\frac{dV}{dt}=3$$, that is, the volume of the pile is increasing at 3 in$$^3/$$sec. The radius of the pile is 2 in, and we need to find $$\frac{dr}{dt}$$, how fast the radius is changing when $$r=2$$.

$$3=\frac23\pi\cdot4\cdot\frac{dr}{dt}$$ $$\frac{dr}{dt} = \frac{9\pi}8$$

The radius of the pile is changing at a rate of $$\frac{9\pi}8$$ in/sec.